Wizard
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Wizard
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March 11th, 2020 at 6:41:01 PM permalink
You have 10 light bulbs going. Each has an average life of 1 year. The life expectancy of each bulb follows an exponential distribution. What this means is the bulbs are never overdue to burn out. Regardless of how long a bulb has been burning, it is no closer to burning out. The average remaining time is always one year per bulb. It is the same principle as waiting for a royal flush in video poker -- no matter how long you've been waiting for one, you're no closer to getting one (assuming no change in strategy).

That said, what is the mean waiting time for the LAST bulb to burn out?



Usual rules:

  1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
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  3. Beer to the first satisfactory answer and solution, subject to rule 2.
  4. Please put answers and solutions in spoiler tags.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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March 12th, 2020 at 3:56:38 PM permalink
Tic Toc. There's still team to join the prestigious beer club. I have had correct answers by PM from those already on it. This puzzle is pretty easy, I think, so if you've ever wanted to gain some math geek points here at WoV, this is your shot.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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March 14th, 2020 at 10:20:11 AM permalink
Any chance of an answer to this soon? I can't seem to find a function of probability of any sort versus time that works.
LoquaciousMoFW
LoquaciousMoFW
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March 14th, 2020 at 3:25:43 PM permalink
I know better than to attempt math/probability problems (better to be thought a fool than speak up and remove all doubt), but I might have had a light turn on, so:
1/365 chance of a bulb burning out on that day, but 10 bulbs to start
so, expected time to lose first bulb is 10/365 or 1/36.5 per day, or 36.5 days average time to lose first bulb.
9 bulbs left, 9/365 chance per day (365/9=40.5556) so 40.5556 days for next bulb (77.056 days total)
8/365 chance per day (365/8=45.625) 122.681 total
7/365=52.1429 122.681+52.1429=174.8239 days so far
6/365; 365/6=60.8333; 60.8333+174.8239= 235.6572
5/365; 365/5=73; 73+235.6572=308.6572
4/365; 365/4=91.25; 91.25+308.6572=399.9072
3/365; 365/3=121.6667; 121.6667+399.9072=521.5739
2/365; 365/2=182.5; 182.5+399.9072=582.4072
1/365; 365/1=365; 365+582.4072=947.4072 days expected until all 10 burn out. (2.5956 years)
Wizard
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Wizard
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March 14th, 2020 at 5:18:50 PM permalink
This one is wide open to members and non-members of the beer club. I would hate to have to post a solution to this fairly easy problem. Very basic probability and statistics.
It's not whether you win or lose; it's whether or not you had a good bet.
ksdjdj
ksdjdj
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March 14th, 2020 at 5:38:37 PM permalink
Quote: Wizard

This one is wide open to members and non-members of the beer club. I would hate to have to post a solution to this fairly easy problem. Very basic probability and statistics.


Thanks Wiz, for making "us simpletons*** " feel even worse.

Is the solution that "easy", that it will probably be an "aha" or "light-bulb" moment for most of us that didn't think of it in time?

***: I am a "statistics simpleton", yet I also consider myself "above average at probability".
Wizard
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Wizard
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March 14th, 2020 at 6:35:12 PM permalink
Quote: ksdjdj

Thanks Wiz, for making "us simpletons*** " feel even worse.



I'm sorry. That could have been worded better.

I will say one probably needs to know the basics of the exponential distribution to solve this one. If you don't, I would leave this one alone.
It's not whether you win or lose; it's whether or not you had a good bet.
ksdjdj
ksdjdj
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March 14th, 2020 at 8:56:46 PM permalink
Quote: Wizard

(snip) I will say one probably needs to know the basics of the exponential distribution to solve this one. If you don't, I would leave this one alone.


I will probably leave this one alone, but do we need to work out the standard deviation to solve this puzzle and if yes is it (see spoiler)
Does the standard deviation equal to the "average life" (mean)?
Wizard
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Wizard
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Thanks for this post from:
ksdjdj
March 15th, 2020 at 4:49:55 AM permalink
Quote: ksdjdj

Does the standard deviation equal to the "average life" (mean)?



Yes
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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March 15th, 2020 at 4:29:06 PM permalink
Unless there is a request for more time, I will post the answer and solution myself tomorrow.
It's not whether you win or lose; it's whether or not you had a good bet.

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